Difference between revisions of "2018 AMC 10B Problems/Problem 17"
Lakecomo224 (talk | contribs) (Created page with "In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and...") |
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+ | == Problem == | ||
+ | |||
In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, points <math>E</math> and <math>F</math> lie on <math>\overline{RS}</math>, and points <math>G</math> and <math>H</math> lie on <math>\overline{SP}</math> so that <math>AP=BQ<4</math> and the convex octagon <math>ABCDEFGH</math> is equilateral. The length of a side of this octagon can be expressed in the form <math>k+m\sqrt{n}</math>, where <math>k</math>, <math>m</math>, and <math>n</math> are integers and <math>n</math> is not divisible by the square of any prime. What is <math>k+m+n</math>? | In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, points <math>E</math> and <math>F</math> lie on <math>\overline{RS}</math>, and points <math>G</math> and <math>H</math> lie on <math>\overline{SP}</math> so that <math>AP=BQ<4</math> and the convex octagon <math>ABCDEFGH</math> is equilateral. The length of a side of this octagon can be expressed in the form <math>k+m\sqrt{n}</math>, where <math>k</math>, <math>m</math>, and <math>n</math> are integers and <math>n</math> is not divisible by the square of any prime. What is <math>k+m+n</math>? | ||
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math> | ||
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+ | == Solution == | ||
+ | |||
+ | Let <math>AP=BQ=x</math>. Then <math>AB=8-2x</math>. | ||
+ | |||
+ | Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>. | ||
+ | |||
+ | Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math>. | ||
+ | |||
+ | Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42) | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=B|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Revision as of 15:03, 16 February 2018
Problem
In rectangle , and . Points and lie on , points and lie on , points and lie on , and points and lie on so that and the convex octagon is equilateral. The length of a side of this octagon can be expressed in the form , where , , and are integers and is not divisible by the square of any prime. What is ?
Solution
Let . Then .
Now notice that since we have .
Thus by the Pythagorean Theorem we have which becomes .
Our answer is . (Mudkipswims42)
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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